Question

помогите с алгеброй пожалуйста​

Answer 1

Ответ:

$x^{2} -4=x^{2} -2^{2}=(x-2)(x+2)$
$25-9a^{2} =5^{2}-3^{2}a^{2}=5^{2}-(3a)^{2}=(5-3a)(5+3a)$
$36m^{2}-100n^{2}=4(9m^{2}-25n^{2})=4(3m-5n)(3m+5n)$$0,04p^{2}-1,69q^{2}=\frac{1}{25}p^{2}-\frac{169}{100}q^{2}=\frac{1}{100}(4p^{2}-169q^{2}) = 0,01(2p-13q)(2p+13q)$$x^{2} y^{2}-\frac{4}{9}=(xy)^{2} -(\frac{2}{3})^{2}=(xy-\frac{2}{3})(xy+\frac{2}{3})$
$a^{4}-b^{6}=(a^{2})^{2}-(b^{3})^{2}=(a^{2}-b^{3})(a^{2}+b^{3})$
$0,01c^{2}-d^{8} =\frac{1}{100}c^{2}-d^{8}=(0,1-d^{4})(0,1c+d^{4})$
$0,81y^{10}-400z^{12} =\frac{81}{100}y^{10}-400z^{12}=(0,9y^{5}-20z^{6})(0,9y^{5}+20z^{6})$
$-1+49a^{4}b^{8}=49a^{4}b^{8}-1=(7a^{2}b^{4}-1)(7a^{2}b^{4}+1)$
$1\frac{7}{9}m^{2}n^{2}-1\frac{11}{25}a^{6}b^{2}=\frac{16}{9}m^{2}n^{2}-\frac{36}{25}a^{6}b^{2}=\frac{4}{225}(100m^{2}n^{2}-81a^{6}b^{2})=0,01(10mn-9a^{3}b)(10mn+9a^{3}b)$